What is the extreme f(x)= x/(4+x^2) ?

The extreme f(x)= x/(4+x^2) is Minimum(-2,-1/4),Maximum(2, 1/4)

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extreme f(x)= x/(4+x^2)

extreme points

f (x) = \frac{x}{4 + x ^{2}}

Minimum (- 2, - \frac{1}{4}), Maximum (2, \frac{1}{4})

Solution steps

f^{'} (x) = \frac{4 - x ^{2}}{( 4 + x ^{2} ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{x}{4 + x ^{2}} : - \frac{1}{4}

Minimum (- 2, - \frac{1}{4})

Plug

x = 2

into

\frac{x}{4 + x ^{2}} : \frac{1}{4}

Maximum (2, \frac{1}{4})

Minimum (- 2, - \frac{1}{4}), Maximum (2, \frac{1}{4})

f (x) = x^{4} - 2 x^{2} + 1

f (x) = \frac{( x ^{2} + 4 )}{x ^{2} - 9}

sec (x) csc (x)

f (x) = 3 x - x^{3}

f (x) = lo g_{6} (x + 7)

What is the extreme f(x)= x/(4+x^2) ?
The extreme f(x)= x/(4+x^2) is Minimum(-2,-1/4),Maximum(2, 1/4)